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Old August 8th, 2009, 05:30 PM
luobo luobo is offline
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This is a pretty simple structural engineering problem. The problem is static determinate.

Note the left support of the beam is a pin and the right is a roller, which means there is no axial force in the beam (i.e., N=0)and there is no moment at both of its ends. There is only shear force at the ends.

By symmetry, there is no shear force at the middle of the beam (V=0).

Hence, there is only moment at the middle of the beam.

Denote the peak of each triangular force by q, each concentrated force by Q, the total span of the beam by L, respectively, then the restraint force at either support is
R=\frac{1}{4}qL+Q

The moment at the middle of the beam is then
M=R\times\frac{L}{2}-\frac{1}{4}qL\times\frac{L}{6}-Q\times\frac{1}{4}L=\frac{1}{12}qL^2+\frac{1}{4}QL

Luobo

Last edited by mr fantastic; September 19th, 2009 at 02:54 AM. Reason: Restored original reply
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